CJC For Sharing 2012 JC2 H2 Prelim Paper 1 Solutions

CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 9740 JC2 PRELIM PAPER 1 SOLUTIONS 2012 1 INEQUALITIES + COMPLEX NUMBERS Part Assessment Objectives Mark Scheme Feedback (i) - solve an equality involving modulus functions either algebraically or graphically - deduce the solution of an inequality of the form 0)( )( xg xf where f(x) and g(x) are linear expressions, based on the solution obtained from solving the equality Suppose ,w       20 023 063 12144 112 112 2 22 22       w ww ww wwww ww ww   1,2,0 023 112 11 12      www ww ww w w (ii) - express a complex number in its algebraic form - represent complex numbers expressed in cartesian form by points in the Argand diagram Suppose ,Cw let iyxw  , where x, y . Since 112  ww ,   . 02 , 0363 , 124144 , || that note , )1(412 , )1(2)12( , )1(2)12( , 11)(2 2222 2222 2222222 22       yxxyxx yxxyxx babiayxyx iyxyix iyxyix iyxiyx

CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 9740 JC2 PRELIM PAPER 1 SOLUTIONS 2012 1 INEQUALITIES + COMPLEX NUMBERS Part Assessment Objectives Mark Scheme Feedback Then, via completing the square,   11 22  yx , which represents a circle of radius 1 and centred about (1, 0), together with its interior.

CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 9740 JC2 PRELIM PAPER 1 SOLUTIONS 2012 2 AP & GP Part Assessment Objectives Mark Scheme Feedback (a) Let n be the no. of months. Then 2000)10)1(100(2 2000))1(2(2   nn dnan 2000)1090(2  nn 040092  nn Solving, 16,25  nn . 16 complete months to fully repay the debt. (b)(i) Find the general of a GP and use it to find a particular value Balance at end of 1 month Balance at end of 2 months [ ] Balance at end of n months ( ) Outstanding amount at 1 January 2013 (ii) Solve inequality using general term. loan repaid ⇒ ⇒ ⇒ so he takes 78 months.

CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 9740 JC2 PRELIM PAPER 1 SOLUTIONS 2012 3 SIGMA NOTATION/METHOD OF DIFF/MI Part Assessment Objectives Mark Scheme Feedback (i) Able to prove a statement using the method of mathematical induction involving a recurrence relation. Let nP be the statement   2 1 1   n nun for all 1n . When 1n , LHS 01  u RHS   0 11 11 2    Since LHS = RHS, 1P is true and forms the basis for induction. Assume kP to be true, i.e.   2 1 1   k kuk . RTP:   21 2  k kuk . LHS     2 2 1 21 4    kk kkuu kk       2 2 2 21 4 1 1     kk kk k k                      22 2 22 23 22 223 22 22 21 12 21 2 21 443 21 421         kk kkk kk kkk kk kkkk kk kkkk   RHS 2 2    k k Therefore, 1 kk PP . Hence by MI, nP is true for all 1n .

CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 9740 JC2 PRELIM PAPER 1 SOLUTIONS 2012 3 SIGMA NOTATION/METHOD OF DIFF/MI Part Assessment Objectives Mark Scheme Feedback (ii) Able to apply the method of difference to find the sum of a series. Able to evaluate sum to infinity of a convergent series.            N n nn N n uu nn nn 2 1 2 2 2 21 4     22 21 12 12 11 11       N N uuN   9 1 2 2    N N Sum to infinity 9 1 (iii) HOT Able to identify a change of index for a summation in order to evaluate it using a previously known summation.                    2 2 2 2 2 2 2111 114 1 32 nn nn nn nn nn         1 2 2 21 4 r rr rr     9 1 2111 114 2 2    0 4 VECTORS Part Assessment Objectives Mark Scheme Feedback

CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 9740 JC2 PRELIM PAPER 1 SOLUTIONS 2012 4 VECTORS Part Assessment Objectives Mark Scheme Feedback (i) Determine if 2 lines intersect and find intersection point.                           , 1 0 2 1 2 1 2,12 1 ryzx , and                           , 1 1 1 2 4 3 243 rzyx . At the intersection point (denoted by P),                                 2 4 3 1 2 21 OP , for some scalars λ, μ. Solving, 2,1   . (via our GC.) Hence,                          0 2 1 1 2 21 1  OP . As such, l1 and l2 intersect at the point P ≡ (-1, 2, 0). Alternative solution : Let ),,( zyxP denote the point of intersection between lines l1 and l2, if it exists. Since P lies on l1, 12 1  zx and 2y .

CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 9740 JC2 PRELIM PAPER 1 SOLUTIONS 2012 4 VECTORS Part Assessment Objectives Mark Scheme Feedback Since P lies on l2, 243  zyx . As 2y and yx  43 in particular, 1x . As 2y and 24  zy in particular, 0z . Check that )0 ,2 ,1() , ,( zyxP satisfies the remaining equation 12 1  zx --- (eq. 1), so as to determine whether )0 ,2 ,1() , ,( zyxP is indeed a point of intersection btw. the lines l1 and l2, i.e. satisfies all equations of line l1 as well as all equations of line l2. LHS of (eq. 1) = 12 11  . RHS of (eq. 1) = 110  . As such the lines l1 and l2 intersect at )0 ,2 ,1( . (ii) Find angle between 2 lines The acute angle between lines l1 and l2 is , 35 |1|cos 1 1 1 1 0 2 1 1 1 1 0 2 cos 11                                                                       . , ...309638.1or ...03678.75 

CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 9740 JC2 PRELIM PAPER 1 SOLUTIONS 2012 4 VECTORS Part Assessment Objectives Mark Scheme Feedback  0.75 (1 d.p.) or 31.1 (3 s.f.). (Accept answer for angle in radians, which is dimensionless.) (iii) Find foot of perpendicular from point to line Use Midpoint/Ratio Theorem to find the reflection point Form equation of line joining 2 points given their position vectors Let N be foot of perpendicular from A(1, 2, -1) to l2, and A’ the reflection of A in l2. (Note that the point (-1, 2, 0) lies on line l3.) Since 2lN                  2 4 3 ON for some  . Since 2lAN  ,

CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 9740 JC2 PRELIM PAPER 1 SOLUTIONS 2012 4 VECTORS Part Assessment Objectives Mark Scheme Feedback so  0 1 1 1 2for vector direction a             l AN , i.e. 0 1 1 1 1 2 4                             . 3 7 . As such,                           1 5 2 3 1 2 4 3 3 7   ON .            5 4 7 3 12' )'(2 1 OAONOAOAOAON A direction vector of l3 is                                  5 2 4 3 1 5 4 7 3 1 0 2 1 . So an equation of l3 is                          , 5 2 4 0 2 1 r . Alternative Solution 1 : Let 3m be a direction vector for line l3.

CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 9740 JC2 PRELIM PAPER 1 SOLUTIONS 2012 4 VECTORS Part Assessment Objectives Mark Scheme Feedback Since the acute angle between l2 and l3 is the same as the acute angle between l1 and l2, which is )(cos 15 11 so ||||15 1 32 32 mm mm  (eq. 1) Since the (obtuse) angle between l1 and l3 is twice the acute angle between l1 and l2, which is )(cos 15 11 so ||||15 1cos2cos 31 31 .15 13115 2 1 15 1coscos2 1 2 1 mm mm                                      (eq. 2) Choose 3m to be a unit vector, i.e. 1 || 3 m (eq. 3)